Unfortunately, I don't know what you have to work with. If Ax= 0 has more than one solution, then the kernel of A is non-trivial and has non-zero dimension. That, in turn, means that the image of A is not all of Rn.

Note also that is Ax= b and v is in the kernel of A, A(x+ v)= Ax+ Av= b+ 0= b.

Ax=0
if you solve for x then x=0,
which will make A invertible
So, what does that say about Ax=B

If x= 0 is the only solution to Ax= 0, then A is invertible. If A has an inverse, how would you solve Ax= B?

However, that is completely irrelevant to your question since the question specifically says that A has MORE than one solution.

I can only repeat: if v is any vector such that Av= 0, and x is a solution to Ax= b, then A(x+ v)= Ax+ Av= Ax+ 0= Ax= b. If there are many solutions to Ax= 0, what does that tell you about the number of solutions to Ax=b?